In Atiyah/Macdonald's Introduction to Commutative Algebra, the following proposition is given (4.7):
Let $\mathfrak{a}$ be a decomposable ideal in a ring $A$, let $\mathfrak{a} = \bigcap_{i=1}^n\mathfrak{q}_i$ be a minimal primary decomposition, and let $r(\mathfrak{q}_i) = \mathfrak{p}_i$. Then $$ \bigcup_{i=1}^n \mathfrak{p}_i = \{x \in A : (\mathfrak{a}:x) \neq \mathfrak{a}\}. $$ In particular, if the zero ideal is decomposable, the set $D$ of zero-divisors of $A$ is the union of the prime ideals belonging to $0$.
The proof begins by noting that we can reduce to just proving the last sentence without loss of generality, because if $\mathfrak{a} = \bigcap_{i=1}^n\mathfrak{q}_i$, then in $A/\mathfrak{a}$, $0 = \bigcap_{i=1}^n\bar{\mathfrak{q}}_i$, and each $\bar{\mathfrak{q}}_i$ is primary, so $0$ is decomposable in the quotient. There are still some details missing here--to really be able to prove the first part given the last sentence, we also have to know how the primes belonging to $0$ in the quotient relate to the $\mathfrak{p}_i$ belonging to $\mathfrak{a}$ in the original ring. I want to say the primes belonging to $0$ in the quotient are just $\bar{\mathfrak{p}}_i$ but I'm not sure; seems like this involves some annoying checks about how radicals play with quotients. Am I on the right track?
